# Free fall with resistance

1. Oct 15, 2005

### eku_girl83

Here's the problem:
Consider a particle of mass m whose motion starts from rest in a constant gravitational field. If a resisting force proportional to the square of the velocity (kmv2) is encountered, find the distance s the particle falls in accelerating from v0 to v1.

I began with the equation m (dv/dt)=-mg-kmv2.
From this, we see that dv = -g-kv2 dt.
When I separate variables and integrate this however, I get a function that involves the inverse tangent. I would have to integrate a second time to find distance, correct?
The book gives the distance s from v0 to v1 as 1/2k ln ((g-kv02)/(g-kv12)). What am I doing wrong???

Any help greatly appreciated!!!

2. Oct 15, 2005

### sqrt(-1)

Use the relation

$$\ddot{x} = v \frac{dv}{dx}$$

on the left hand side of your equation and you should find things work out :)